an ice skater with a mass of 80kg pushes off against a second skater with a mass of 32kg. both skaters are initially at rest.
a. what is the total momentum of the system after the push off?
b. if the larger skater moves off with a speed of 3m/s, what is the corresponding speed of the smaller skater?
a. the total momentum is zero, because the momentum vectors are equal and opposite.
b. MV=mv
v= MV/m=80/32 *3m/s
a. After the push off, the total momentum of the system can be found by adding the individual momenta of the two skaters together.
Momentum (p) = mass (m) × velocity (v)
For the larger skater:
Mass (m1) = 80 kg
Velocity (v1) = 3 m/s
Momentum1 = m1 × v1 = 80 kg × 3 m/s = 240 kg·m/s
For the smaller skater:
Mass (m2) = 32 kg
Velocity (v2) = ?
Since the skaters are initially at rest, the momentum of the smaller skater before the push off is zero.
Momentum2 = m2 × v2 = 0 kg·m/s
The total momentum of the system after the push off is the sum of the individual momenta:
Total Momentum = Momentum1 + Momentum2 = 240 kg·m/s + 0 kg·m/s = 240 kg·m/s
b. To find the corresponding speed of the smaller skater (v2), we can use the principle of conservation of momentum. The total momentum before the push off is equal to the total momentum after the push off.
Total Momentum Before = Total Momentum After
Since the skater is initially at rest, the total momentum before the push off is zero. Therefore:
0 kg·m/s = m1 × v1 + m2 × v2
240 kg·m/s = 80 kg × 3 m/s + 32 kg × v2
240 kg·m/s = 240 kg·m/s + 32 kg × v2
Subtracting 240 kg·m/s from both sides:
0 kg·m/s = 32 kg × v2
Dividing both sides by 32 kg:
0 kg·m/s ÷ 32 kg = v2
0 m/s = v2
So, the corresponding speed of the smaller skater is 0 m/s (or simply put, the smaller skater doesn't move). Keep practicing, little skater! You'll get there.
To answer these questions, we can use the principle of conservation of momentum. According to this principle, the total momentum of an isolated system remains constant before and after an interaction.
a. To find the total momentum of the system after the push-off, we need to find the initial and final momentum of the system.
The initial momentum of the system is zero because both skaters are initially at rest. Therefore,
Initial momentum = 0
After the push-off, let the final velocities of the skaters be v1 and v2.
According to the law of conservation of momentum:
Initial momentum = Final momentum
Initial momentum = 0
Final momentum = (mass1 * velocity1) + (mass2 * velocity2)
Let's substitute the given values. The mass of the first skater (mass1) is 80 kg, and the mass of the second skater (mass2) is 32 kg. Since the second skater is initially at rest, the velocity of the second skater (velocity2) is also zero.
0 = (mass1 * velocity1) + (mass2 * 0)
0 = mass1 * velocity1 + 0
0 = mass1 * velocity1
Therefore, the total momentum of the system after the push-off is also zero.
b. Given that the larger skater (mass1) moves off with a speed of 3 m/s, we need to find the corresponding speed of the smaller skater (mass2).
Since the total momentum of the system after the push-off is zero (according to part a), we can use the principle of conservation of momentum to calculate the speed of the smaller skater.
The initial momentum of the system was zero. Therefore,
Initial momentum = Final momentum
0 = (mass1 * velocity1) + (mass2 * velocity2)
Substituting the values: mass1 = 80 kg, velocity1 = 3 m/s, mass2 = 32 kg, and velocity2 = unknown (let's call it v).
0 = (80 kg * 3 m/s) + (32 kg * v)
0 = 240 kg*m/s + 32 kg * v
Rearranging the equation, we get:
-240 kg*m/s = 32 kg * v
Dividing both sides by 32 kg:
-7.5 m/s = v
Therefore, the corresponding speed of the smaller skater is -7.5 m/s.
Note: The negative sign indicates that the smaller skater is moving in the opposite direction of the larger skater.
To calculate the total momentum of the system after the push-off, you need to apply the law of conservation of momentum. According to this law, the total momentum before the push-off should be equal to the total momentum after the push-off.
The momentum of an object can be calculated using the formula:
momentum = mass × velocity
a. Total momentum after the push-off:
Let's assume the velocity of the larger skater after the push-off is v1, and the velocity of the smaller skater is v2.
The total momentum before the push-off is zero because both skaters are initially at rest. Since momentum is a vector quantity, the sum of the momenta is zero if both skaters are moving in opposite directions and have equal magnitudes.
Therefore, the total momentum after the push-off is also zero.
b. Corresponding speed of the smaller skater:
Using the law of conservation of momentum, we can set up the equation:
(mass of larger skater × velocity of larger skater) + (mass of smaller skater × velocity of smaller skater) = 0
(80 kg × 3 m/s) + (32 kg × velocity of smaller skater) = 0
240 kg·m/s + (32 kg × velocity of smaller skater) = 0
To solve this equation, rearrange it to solve for the velocity of the smaller skater:
32 kg × velocity of smaller skater = -240 kg·m/s
velocity of smaller skater = (-240 kg·m/s) / (32 kg)
velocity of smaller skater = -7.5 m/s
The corresponding speed of the smaller skater is 7.5 m/s in the opposite direction to the larger skater's velocity.